Odds of someone reshoving over our shove given their ranges

[djk123]

Say we shove over a short stack, and there are 3 people left to act. Their ranges are 77+,ATs+,KQs,AJo+,KQo, TT+,AQ+, and 77+,ATs+,KQs,AJo+,KQo. So the first and third guy will have one of those hands 8.7% of the time, and the 2nd guy will have one of those hands 4.7% of the time.

How do we calculate the odds we get called by one of them? Say if the first guy shoves, and the other 2 guys behind will only come in with QQ+ or KK+, that situation is so unlikely we can basically ignore it right?

[djk123]

nvm i figured it out

[jay_shark]

Show us that you figured it out . You'd be surprised how many mistakes people make in situations like this .

To give an accurate response , it is always good to know how short your short stack is . There can be a difference in shoving with 8BB or 10BB so it's important you give this information .

[djk123]

Well say their ranges are 77+,ATs+,AJo+,KQ ; TT+,AQ+ ; 77+,ATs+,AJo+,KQ. First and third are 8.7% of hands. 2nd one is 4.7% of hands. Also, say that if first guy shoves, 2nd and 3rd guy wud only come in with QQ+, and if the first 2 guy were to shove the third guy would only come in with KK+.

Chance that all fold would be (1-.087)* (1-.047) * (1-.087), right?

Chance that first guy shoves and other two fold : (.087) * (1-.014) * (1-.014)

and so on for the 6 other possibilites. I think I did it right because i worked the chances out for all 8 possible outcomes and the probabilities added up to 1.

After we shove, there are 8 possible outcomes: P being push, F being fold.

P F F (.084581) ; P P F (.001207) ; P F P (.001201) ;

P P P (.000011) ; F P F (.04231) ; F F P (.075698) ;

F P P (.000601) ; F F F (.794391)

[sickofants]

It looks like your maths is good there. Just as a quick aside, you haven't taken into account what hand you are pushing with.

For instance, if you're pushing with, say Ah6c then you can take those cards out of the deck and calculate:

Pr(Player 2 has hand within his range) = 51/1225 = 4.2%
Pr(Within 1 or 3's range) = 97/1225 = 7.9%

so we have

Pr(F F F) = 81%

Unless I've made some stupid counting error, of course.